diff options
Diffstat (limited to '249/CH3/EX3.2/3_02.sce')
-rwxr-xr-x | 249/CH3/EX3.2/3_02.sce | 63 |
1 files changed, 37 insertions, 26 deletions
diff --git a/249/CH3/EX3.2/3_02.sce b/249/CH3/EX3.2/3_02.sce index a2a1ef1b6..bb129725c 100755 --- a/249/CH3/EX3.2/3_02.sce +++ b/249/CH3/EX3.2/3_02.sce @@ -1,26 +1,37 @@ -clear
-clc
-CA=[10;8;6;5;3;2;1];//mol/litre
-T=[0;20;40;60;120;180;300];//sec
-//plot(T,CA)
-//xlabel('Time(sec)');ylabel('CA(mol/litre)');
-//From graph y=-dCA/dt at different points are
-y=[-0.1333;-0.1031;-0.0658;-0.0410;-0.0238;-0.0108;-0.0065];
-//Guessing nth rate order
-//rA=kCA^n
-//log(-dCA/dt)=logk+nlogCA
-for i=1:7
-log_y(i)=log10(y(i));
-log_CA(i)=log10(CA(i));
-end
-plot(log_CA,log_y)
-xlabel('logCA');ylabel('log(-dCA/dt)')
-coeff1=regress(log_CA,log_y);
-n=coeff1(2);
-k=-10^(coeff1(1));
-printf("\n After doing linear regression,the slope and intercept of the graph is %f , %f",coeff(2),coeff(1))
-printf("\n The rate equation is therefore given by %f",k)
-printf("CA^1.375 mol/litre.sec")
-disp('The answer slightly differs from those given in book as regress fn is used for calculating slope and intercept')
-
-
+clear +clc +function [coefs]=regress(x,y) +coefs=[] + if (type(x) <> 1)|(type(y)<>1) then error(msprintf(gettext("%s: Wrong type for input arguments: Numerical expected.\n"),"regress")), end + lx=length(x) + if lx<>length(y) then error(msprintf(gettext("%s: Wrong size for both input arguments: same size expected.\n"),"regress")), end + if lx==0 then error(msprintf(gettext("%s: Wrong size for input argument #%d: Must be > %d.\n"),"regress", 1, 0)), end + x=matrix(x,lx,1) + y=matrix(y,lx,1) + xbar=sum(x)/lx + ybar=sum(y)/lx + coefs(2)=sum((x-xbar).*(y-ybar))/sum((x-xbar).^2) + coefs(1)=ybar-coefs(2)*xbar +endfunction +CA=[10;8;6;5;3;2;1];//mol/litre +T=[0;20;40;60;120;180;300];//sec +//plot(T,CA) +//xlabel('Time(sec)');ylabel('CA(mol/litre)'); +//From graph y=-dCA/dt at different points are +y=[-0.1333;-0.1031;-0.0658;-0.0410;-0.0238;-0.0108;-0.0065]; +//Guessing nth rate order +//rA=kCA^n +//log(-dCA/dt)=logk+nlogCA +for i=1:7 +log_y(i)=log10(y(i)); +log_CA(i)=log10(CA(i)); +end +plot(log_CA,log_y) +xlabel('logCA');ylabel('log(-dCA/dt)') +coeff1=regress(log_CA,log_y); +n=coeff1(2); +k=-10^(coeff1(1)); +printf("\n After doing linear regression,the slope and intercept of the graph is %f , %f",coeff(2),coeff(1)) +printf("\n The rate equation is therefore given by %f",k) +printf("CA^1.375 mol/litre.sec") +disp('The answer slightly differs from those given in book as regress fn is used for calculating slope and intercept')
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